我对最佳方法有疑问。当数据被认为是可变的时候,我不确定哪种方法最好。
考虑以下3个表:
员工
EMPLOYEE_ID,EMP_NAME
PROJECT
PROJECT_ID,PROJ_NAME
EMP_PROJ (上述两个表中有多对多)
EMPLOYEE_ID,PROJECT_ID
问题:在给定EmployeeID的情况下,找到与此员工关联的所有项目的所有员工。
我以两种方式尝试过这种方法。无论使用何种大小的数据,这两种方法的差别仅为几毫秒。
SELECT EMP_NAME FROM EMPLOYEE
WHERE EMPLOYEE_ID IN (
SELECT EMPLOYEE_ID FROM EMP_PROJ
WHERE PROJECT_ID IN (
SELECT PROJECT_ID FROM EMP_PROJ p, EMPLOYEE e
WHERE p.EMPLOYEE_ID = E.EMPLOYEE_ID
AND E.EMPLOYEE_ID = 123)
去
select c.EMP_NAME FROM
(SELECT PROJECT_ID FROM EMP_PROJ
WHERE EMPLOYEE_ID = 123) a
JOIN
EMP_PROJ b
ON a.PROJECT_ID = b.PROJECT_ID
JOIN
EMPLOYEE c
ON b.EMPLOYEE_ID = c.EMPLOYEE_ID
截至目前,我预计每个人约有5000名员工和项目......但不知道存在多少种关系。 你会推荐哪种方法? 谢谢!
编辑: 方法1的执行计划
"Hash Join (cost=86.55..106.11 rows=200 width=98)"
" Hash Cond: (employee.employee_id = emp_proj.employee_id)"
" -> Seq Scan on employee (cost=0.00..16.10 rows=610 width=102)"
" -> Hash (cost=85.07..85.07 rows=118 width=4)"
" -> HashAggregate (cost=83.89..85.07 rows=118 width=4)"
" -> Hash Semi Join (cost=45.27..83.60 rows=118 width=4)"
" Hash Cond: (emp_proj.project_id = p.project_id)"
" -> Seq Scan on emp_proj (cost=0.00..31.40 rows=2140 width=8)"
" -> Hash (cost=45.13..45.13 rows=11 width=4)"
" -> Nested Loop (cost=0.00..45.13 rows=11 width=4)"
" -> Index Scan using employee_pkey on employee e (cost=0.00..8.27 rows=1 width=4)"
" Index Cond: (employee_id = 123)"
" -> Seq Scan on emp_proj p (cost=0.00..36.75 rows=11 width=8)"
" Filter: (p.employee_id = 123)"
方法2的执行计划:
"Nested Loop (cost=60.61..112.29 rows=118 width=98)"
" -> Index Scan using employee_pkey on employee e (cost=0.00..8.27 rows=1 width=4)"
" Index Cond: (employee_id = 123)"
" -> Hash Join (cost=60.61..102.84 rows=118 width=102)"
" Hash Cond: (b.employee_id = c.employee_id)"
" -> Hash Join (cost=36.89..77.49 rows=118 width=8)"
" Hash Cond: (b.project_id = p.project_id)"
" -> Seq Scan on emp_proj b (cost=0.00..31.40 rows=2140 width=8)"
" -> Hash (cost=36.75..36.75 rows=11 width=8)"
" -> Seq Scan on emp_proj p (cost=0.00..36.75 rows=11 width=8)"
" Filter: (employee_id = 123)"
" -> Hash (cost=16.10..16.10 rows=610 width=102)"
" -> Seq Scan on employee c (cost=0.00..16.10 rows=610 width=102)"
所以看起来方法2的执行计划略好一些,因为'成本'是60而不是方法1的85.这是分析这个的正确方法吗?
即使对于各种多种组合,人们如何知道它仍然适用?
答案 0 :(得分:4)
两种方法可能导致次优性能,因为它们都使用根据DBMS有效优化或不优化的子查询。
我建议避免使用子查询,并使用JOIN
来表示要使用的内容,例如:
编辑:
我纠正并简化了我的例子:
select coll.EMP_NAME
from EMP_PROJ ep1
inner join EMP_PROJ ep2 on ep1.PROJECT_ID = ep2.PROJECT_ID
inner join EMPLOYEE coll on ep2.EMPLOYEE_ID = coll.EMPLOYEE_ID
where ep1.EMPLOYEE_ID = 123
我认为应该注意,在一个查询中使用不同的别名多次引用相同的表是完全合法的。对于查询,这在逻辑上“看起来”就像两个具有相同结构和数据的单独表。